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For the following geometric series

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For $\\sum\\limits_{i=0}^\\var{upsum} \\frac{1}{\\var{mult2}} \\times (\\var{r2})^{i}$

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What is the first term? [[0]]

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What is the common ratio? [[1]]

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What is the value of the sum? [[2]]

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We will look at solutions for a). The method for b) is exactly the same.

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To find the first term, plug in i=0. That is $\\frac{1}{\\var{mult2}} \\times (\\var{r2})^{0}=\\frac{1}{\\var{mult2}}\\times 1 = \\frac{1}{\\var{mult2}}$

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To find the common ratio, we look at the $(i+1)$th term divided by the $i$th term. That is $\\frac{\\frac{1}{\\var{mult2}}\\times \\simplify{{r2}}^{i}}{\\frac{1}{\\var{mult2}}\\times \\simplify{{r2}}^{i-1}} = \\var{r2}$

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To find the value of the sum, we apply the known formula for a geometric series for finding the sum of the first $n$ terms. That is $S_n = \\frac{a_1(1-r^n)}{1-r}$, where $a_1$ is the first term and $r$ is the common ratio. In our case we have $S_n = \\frac{\\frac{1}{\\var{mult2}}(1-\\var{r2}^n)}{1-\\var{r2}}$

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